Numerical Analysis 6.1. Lagrange Interpolation скачать в хорошем качестве

Numerical Analysis 6.1. Lagrange Interpolation

This video is a comprehensive lecture on Numerical Analysis, specifically focusing on Lagrange Interpolation. Below is a summary of the key concepts discussed: 1. The Lagrange Interpolation Problem The core goal is to find a function G from a specific class (polynomials) that passes through a set of given points (x 0 ​ ,y 0 ​ ),(x 1 ​ ,y 1 ​ ),…,(x n ​ ,y n ​ ) where the x values are distinct mesh points [00:27]. Definition: We seek a polynomial L n ​ of degree at most n such that L n ​ (x i ​ )=y i ​ for all i from 0 to n [01:41]. Uniqueness: The lecturer proves that this problem has a unique solution using a proof by contradiction, showing that the difference between two such polynomials would have more roots than its degree allows, meaning the polynomials must be identical [10:07]. 2. Lagrange Basis Polynomials To build the interpolating polynomial, the lecture introduces Lagrange basis polynomials (ℓ k ​ (x)): Construction: Each basis polynomial is a product of linear terms (x−x i ​ ) divided by constants (x k ​ −x i ​ ), skipping the term where i=k to avoid division by zero [02:54]. Property: A basis polynomial ℓ k ​ (x) equals 1 at its corresponding mesh point x k ​ and 0 at all other mesh points x i ​ [06:19]. Final Formula: The complete Lagrange polynomial is the sum of these basis polynomials weighted by the target y values: L n ​ (x)=∑y k ​ ℓ k ​ (x) [06:43]. 3. Error Analysis and Rolle's Theorem The lecture details how to estimate the error between the original function f(x) and the interpolating polynomial L n ​ (x) [27:13]. Generalized Rolle's Theorem: To prove the error formula, the lecturer reviews Rolle's Theorem and its generalization, which states that if a function has n+1 roots, its n-th derivative must have at least one root [22:25]. Error Term: The error f(x)−L n ​ (x) is proportional to the (n+1)-th derivative of f evaluated at some point ξ, divided by (n+1)!, and multiplied by a product of linear terms [28:31]. Convergence: For functions like cos(x), increasing the number of mesh points generally decreases the error within the interval of the mesh points [18:23]. 4. 2D Lagrange Interpolation The lecturer extends the concept to two-dimensional functions defined over a rectangular domain [45:28]. The 2D interpolant is constructed using a double summation of the product of x-direction and y-direction basis polynomials [46:49]. Visual Example: A 3D plot is shown where the resulting surface passes through specific data points (red dots) on a grid [49:18]. Example Provided in the Video A numerical example is worked through at [12:09] using four points to find a cubic polynomial. The lecturer demonstrates how to omit specific points to create the numerator and denominator for each basis term [13:02].